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Applications of Linear Equations

 

Sir William Rowan Hamilton, an Irish mathematician, invented linear equations in the year 1843. He induced relationships between various variables to find their values. To date, we use linear equations to solve numerous mathematical problems.

The general representation of a linear equation is given by: ax + by = c. Here, the variables have degree one, which makes them linear.

Linear equations are used in many scenarios as:

 1. Geometry problems
 2. Age problems
 3. Speed, distance and time problems
 4. Money and finance problems
 5. Ratios and proportions problems
 6. Hourly, daily and monthly wage problems
 7. Pressure and force problems
 8. Area and perimeter problems

Linear equations are widely used to solve relationships between two or more variables that occur in real life. Professionals make relationships between two or more factors and solve those using linear equations and linear algebra.

Here are a few examples based on linear equation applications:

Example: Given that the Sum of Two Numbers is 48. One of those numbers is thrice the other number. What are these numbers?
Solution:

Let the two unknown numbers be x and y.
According to the data given, the sum of both numbers is 48.
Therefore, x + y = 48
Also, one number is thrice the other number,
We will consider y as the bigger number, such that we can write 3x = y.
Using the substitution method and substituting y = 3x from the above equation, we get,
x + 3x = 48
4x = 48
x = 12
Also, y = 3x
Therefore, the value of y is 12 x 3 = 36
y = 36
We can verify it as, x + y = 48
12 + 36 = 48

Since, LHS = RHS, therefore, our solution is correct.

Example 2: What is the speed of a car if it travels from Mumbai to Pune in 4 hours, covering a distance of 800 km?
Solution:

Let the speed of the car be S, time t, and distance d.
We know the relation between time, speed and distance is given by:
Speed = distance/time
S = d / t
Here, d = 800, t = 4.
Therefore, we get,
S = 800/4
S = 200

Hence, the speed of the car is 200 km/hr.

Example 3: At present, the age of the son is one-fourth of his father. Assuming after 5 years, the son becomes one-third of his father’s age, calculate their present ages.
Solution:

Let us consider the present age of the father and son to be x and y, respectively.
According to the data given, the son's present age is one-fourth of his father’s age.
So, y = x/4 or x= 4y
We need to find the age after 5 years, so we add 5 to the present ages.
y + 5) = 1/3 (x + 5)
Or, 3y + 15 = x + 5
Substituting x = 4y from the above equations, we get,
3y + 15 = 4y + 5
4y - 3y = 15 - 5
Hence, y = 10
Now, x = 4y
x = 4 x 10 = 40.
Hence, x = 40.

Therefore, the father's present age is 40 years, and his son’s age is 10 years.

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